The scalar no scalar mystery v2

The scalar no scalar mystery. Here is the discussion on this subject , _______________________________________________________________ Scalar under transformations Imagine a version of Picture A (2.1.1) where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i. Under such a transformation, a dot product like α v transforms as a scalar. So, whereas v and α transform as vectors, the function α(v) transforms as a scalar field, just as in (2.2.6). That is to say, α(v) = α v = α' v' = α'(v') . (2.11.13) Although the function α(v) transforms as a scalar, the associated vector α (covector) transforms as a rank-1 tensor (vector), and it is through this isomorphic connection that we loosely refer to the function α(v) as being a rank-1 tensor. When functional α is expanded on the basis functionals λi, the coefficients αi are the rank-1 tensor components, and these coefficients are projected out by (2.11.9). ____________________________________________________________________ I have just read my Tensor appendix opening section on "scalar vs tensor" and the "three experiments". This is very relevant to the current situation. Go back to this Section 2.11 situation λi(v) = ei v If we do Experiment #1 and we rotate the apparatus and the coordinate system at the same time, we get ei v = e'i v' and we say that λi(v) a scalar. BUT in Experiment #2 we "rotate the apparatus" but keep the coordinate system unchanged, we find λ'i(v') = ei v' = Rijλi(v) This is an elementary vector field transformation rule!!!! My whole "mystery" evaporates. Which kind of transformation is this? I think it is a component transformation!!! Forget that "vector transformation" thing. Now how does this work for the more general case of α instead of λi ? α(v) = α v Write α = Σiαiλi where αi are constants. Then in Experiment #2 we would get α' = Σiαiλ'i and then α'(v') = Σiαiλ'i(v') = Σiαi[ΣjRijλj(v) ] = Σjλj(v) [ΣiαiRij] Now how do you argue that α should not move in Experiment #2? α = Σiαiλi = α λ // this notation is suddenly OK!!! Then for k = 2 I get (λiλj)'(v'1,v'2) = [(ei v1)(ej v2)]' = (ei v'1)(ej v'2) = (v'1)i(v'2)j = RiaRjb(v1)a(v2)b = RiaRjb (λaλb)(v1,v2) so we get a tensor field with two arguments. Question: What happens if x' = F(x) is not linear? Well, the vectors v1 are not themselves functions of space, but they do transform. So would you say T'(v1', v'2.....v'k) = Ria(x)Rjb(x) .....T(v1, v2.....) I write (v'1)i = Rij(x) (v1)j ??? Does this make any sense?? Probably not. Maybe the rule is this T'ab(v1'(x'), v'2(x').....v'k(x')) = Ria(x)Rjb(x) .....Tij(v1(x), v2(x).....) This is a tensor field whose arguments are fields. I think this works! This is a "tensor field example" I did not include in my writeup today. and this is a rank-2 tensor. The whole problem just goes away, there is no scalar involved. I am glad I wrote up that tensor doc thing! Pause. Go back to this, T = Σii....i Tii....i (ei ei ..... ei) . (2.10.14) If you do an Experiment #1 transformation, you get T' = Σii....i T'ii....i (ei ei ..... ei) . and this is a different object, we do not have T' = T. I guess the question is this: is Tii....i part of the apparatus, or is it part of the coordinate system and a constant like π. If I assume Tii....i is the apparatus, like a vector v1 , then we get T'ii....i = RijRij.....Rij Tjj....j Then, T' = Σii....i T'ii....i (ei ei ..... ei) . = Σii....i [RijRij.....Rij Tjj....j] (ei ei ..... ei) = ΣI,J RIJ TJ eI Try a simpler case V = ΣiViei V' = ΣiV'iei // active rotation How then are V and V' related? Express V' in the coordinate system S. V' = Σi [ ΣjRijVj ] ei There is nothing more you can say! So thinking about T' in frame S' is not very helpful. Better to talk about tensors or tensor functions, not abstract tensors.